Question 11 Mark
Assertion (A): Two coins are tossed simultaneously. The probability of getting two heads, if it is known that at least one head comes up, is $\frac{1}{3}$.
Reason $(R)$ : Let $E$ and $F$ be two events with a random experiment, then $P(F / E)=\frac{P(E \cap F)}{P(E)}$.
AnswerSample space $=\{ HH , HT , TH , TT \}$
Let $A$ be the event of coming up two heads
$\therefore \quad A=\{H H\} \Rightarrow P(A)=\frac{1}{4}$
and $B$ be the event of coming up atleast one head
$
\therefore \quad B=\{ HH , HT , TH \} \Rightarrow P(B)=\frac{3}{4}$
Also, $A \cap B=\{H H\} \Rightarrow P(A \cap B)=\frac{1}{4}$
So, required probability $=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
So, assertion is true.
Also, reason is true and it is the correct explanation of assertion.
View full question & answer→Question 21 Mark
Assertion $(A) :$ The probability that candidates $A$ and $B$ can solve the problem is $\frac{1}{5}$ and $\frac{2}{5}$, then probability that problem will be solved is given by $\frac{12}{25}$.
Reason $(R):$ If events $A \& B$ are independent, then $P(A \cap B)=P(A) \times P(B)$.
Answer$(d) :$ Probability of solving the problem by $A \ B$ is $=1-P($ None of them can solve the problem $)$
$=1-P(\bar{A} \cap \bar{B})=1-P(\bar{A}) \cdot P(\bar{B})$
$=1-[1-P(A)][1-P(B)]=1-\frac{4}{5} \times \frac{3}{5}=\frac{13}{25} .$
View full question & answer→Question 31 Mark
Let $A$ and $B$ be two events associated with an experiment such that $P(A \cap B)=P(A) P(B)$.
Assertion (A) : $P(A \mid B)=P(A)$ and $P(B \mid A)=$ $P(B)$
Reason (R): $P(A \cup B)=P(A)+P(B)$.
Answer(c) : Since, $P(A \cap B)=P(A) P(B)$, therefore, $A$ and $B$ are independent events.
$
\therefore \quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A) P(B)}{P(B)}=P(A)
$
Similarly, $P(B \mid A)=P(B)$.
Thus, assertion is true.
However, Reason is not correct for independent events. For example, when a dice is rolled once, then the events $A$ : 'an even number' shows up and $B$ : 'a multiple of 3 ' show up are independent as
$
\begin{aligned}
P(A) P(B)=\frac{3}{6} \times \frac{2}{6}=\frac{1}{6} & =P(A \cap B) \\
& (\because A=\{2,4,6\} \text { and } B=\{3,6\})
\end{aligned}
$
But $P(A \cup B)=P(\{2,3,4,6\})$
$
=\frac{4}{6} \neq P(A)+P(B) \quad\left(\because P(A)+P(B)=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \neq \frac{4}{6}\right)
$
View full question & answer→Question 41 Mark
Assertion (A) : An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. Then, the probability that the second ball is red is $\frac{1}{2}$.
Reason (R) : By Bayes' Theorem
$
\begin{array}{r}
P\left(E_1 \mid E\right)=\frac{P\left(E \mid E_1\right) P\left(E_1\right)}{P\left(E_1\right) P\left(E \mid E_1\right)+P\left(E_2\right) P\left(E \mid E_2\right)+P\left(E_3\right) P\left(E \mid E_3\right)}
\end{array}
$
Answer(a) : Clearly, reason is true. Now, The urn contains 5 red and 5 black balls.
Let $E_1$ : a red ball is drawn in the first attempt and $E_2$ : a black ball is drawn in the first attempt.
Then, $P\left(E_1\right)=P\left(E_2\right)=\frac{5}{10}=\frac{1}{2}$
Now, let $E$ : red ball is drawn in second attempt.
Then, $P\left(E \mid E_1\right)=\frac{7}{12}$ and $P\left(E \mid E_2\right)=\frac{5}{12}$
Now, probability of drawing second ball as red is
$
\begin{aligned}
P(E) & =P\left(E_1\right) \cdot P\left(E \mid E_1\right)+P\left(E_2\right) \cdot P\left(E \mid E_2\right) \\
& =\frac{1}{2} \times \frac{7}{12}+\frac{1}{2} \times \frac{5}{12}=\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)=\frac{1}{2} \times 1=\frac{1}{2}
\end{aligned}
$
$\therefore$ Assertion is true, Reason is true and correct explanation of Assertion.
View full question & answer→Question 51 Mark
Assertion (A) : Bag I contains 3 red and 4 black balls while another bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Then, the probability that it was drawn from bag II is $\frac{35}{68}$.
Reason (R) : By Bayes' theorem,
$
\begin{array}{r}
P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right)\left(A \mid E_1\right)+P\left(E_1\right) \cdot P\left(A \mid E_2\right)+P\left(E_1\right) P\left(A \mid E_3\right)}
\end{array}
$
Answer(c) : Clearly reason is false. Let $E_1$ be the event of choosing the bag $I, E_2$ be the event of choosing the bag II and $A$ be the event of drawing a red ball.
Then, $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$
Also, $P\left(A \mid E_1\right)=P($ drawing a red ball from bag $I)=\frac{3}{7}$
and $P\left(A \mid E_2\right)=P($ drawing a red ball from bag II $)=\frac{5}{11}$
Now, the probability of drawing a ball from bag II, being given that it is red, is $P\left(E_2 \mid A\right)$
By using Bayes' theorem, we have
$
\begin{aligned}
P\left(E_2 \mid A\right) & =\frac{P\left(E_2\right) P\left(A \mid E_2\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)} \\
& =\frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{5}{11}}=\frac{35}{68}
\end{aligned}
$
$\therefore \quad$ Assertion is true but reason is false.
View full question & answer→Question 61 Mark
Assertion $(A) :$ Let $E$ and $F$ be events associated with the sample space $S$ of an experiment. Then, we have $P(S \mid F)=P(F \mid F)=1$.
Reason $(R) :$ If $A$ and $B$ are any two events associated with the sample space $S$ and $F$ is an event associated with $S$ such that $P(F) \neq 0$, then $P((A \cup B) \mid F)=P(A \mid F)+P(B \mid F)-P((A \cap B) \mid F)$
Answer$(b) :$ We know that,
$P(S \mid F)=\frac{P(S \cap F)}{P(F)}=\frac{P(F)}{P(F)}=1 [ F \subset S ]$
Also, $P(F \mid F)=\frac{P(F \cap F)}{P(F)}=\frac{P(F)}{P(F)}=1$
Thus, $P(S \mid F)=P(F \mid F)=1$ So, Assertion is true.
Also, we have,
$P((A \cup B) \mid F)=\frac{P[(A \cup B) \cap F]}{P(F)}=\frac{P[(A \cap F) \cup(B \cap F)]}{P(F)}$
(by distributive law of intersection over union)
$=\frac{P(A \cap F)+P(B \cap F)-P(A \cap B \cap F)}{P(F)}$
$=\frac{P(A \cap F)}{P(F)}+\frac{P(B \cap F)}{P(F)}-\frac{P((A \cap B) \cap F)}{P(F)}$
$=P(A \mid F)+P(B \mid F)-P((A \cap B) \mid F)$
$\therefore $ Reason is true but not the correct explanation of Assertion.
View full question & answer→Question 71 Mark
Consider the experiment of drawing a card from a deck of 52 playing cards, in which the elementary events are assumed to be equally likely.
Assertion (A) : If $E$ and $F$ denote the events the card drawn is a spade and the card drawn is an ace respectively,
then $P(E \mid F)=\frac{1}{4}$ and $P(F \mid E)=\frac{1}{13}$.
Reason (R): $E$ and $F$ are two events such that the probability of occurrence of one of them is not affected by occurrence of the other. Such events are called independent events.
Answer(a): We have, $P(E)=\frac{13}{52}=\frac{1}{4}$ and $P(F)=\frac{4}{52}=\frac{1}{13}$
Also, $E \cap F$ denote the event the card drawn is the ace of spades.
$
\therefore P(E \cap F)=\frac{1}{52}
$
Hence, $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{52}}{\frac{1}{13}}=\frac{1}{4}$
Since, $P(E)=\frac{1}{4}=P(E \mid F)$, we can say that the occurrence of event $F$ has not affected the probability of occurrence of the event $E$. We also have,
$
P(F \mid E)=\frac{P(E \cap F)}{P(E)}=\frac{\frac{1}{52}}{\frac{1}{4}}=\frac{1}{13}=P(F)
$
Again, $P(F)=\frac{1}{13}=P(F \mid E)$ shows that occurrence of event $E$ has not affected the probability of occurrence of the event $F$. Thus, $E$ and $F$ are two events such that the probability of occurrence of one of them is not affected by occurrence of the other.
Such events are called independent events.
View full question & answer→Question 81 Mark
Consider the system of equations $a x+b y=0, c x+d y=0$ where $a, b, c, d \in\{0,1\}$.
Assertion $(A):$ The probability that the system of equations has a unique solution is $\frac{3}{8}$.
Reason $(R):$ The probability that the system of equations has a solution is $1$ .
AnswerThe given system of equations can be written as $A X=O$, where $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$.
As $a, b, c, d \in\{0,1\}$, therefore, matrix $A$ can be chosen in $2^4=16$ ways.
The given system has a unique solution namely the zero solution $($i.e. $x=0, y=0 )$ only if $|A| \neq 0$.
i.e. if $A$ is any one of the following matrices
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]$
$\therefore \ P\ ($ system has a unique solution $) =\frac{6}{16}=\frac{3}{8}$.
Thus, the assertion is true.
Further, we know that a homogeneous system is always consistent and admits of atleast one solution namely the zero solution.
So, the probability that the system has a solution is $1$ .
Thus reason is also true.
View full question & answer→Question 91 Mark
Let $H_1, H_2, \ldots, H_n$ be mutually exclusive and exhaustive events with $P\left(H_i\right) > 0, i=1,2, \ldots, n$. Let $E$ be any other event with $0$ Assertion $(A) : P\left(H_i \mid E\right) > P\left(E \mid H_i\right) P\left(H_i\right)$ for $i=1,2, \ldots, n$
$\text { Reason (R) : } \sum_{i=1}^n P\left(H_i\right)=1 \text {. }$
Answer$\text { (d) }: P\left(H_i \mid E\right) > P\left(E \mid H_i\right) \times P\left(H_i\right)$
$\Rightarrow \frac{P\left(H_i \cap E\right)}{P(E)} > P\left(E \cap H_i\right)$
$\Rightarrow P\left(H_i \cap E\right)(1-P(E)) > 0$
$\Rightarrow P\left(H_i \cap E\right) > 0$
This leads to a contradiction $0 > 0$ if $H_i \cap E=\phi$ for any $i$.
View full question & answer→Question 101 Mark
A man $P$ speaks truth with probability $p$ and an other $\operatorname{man} Q$ speaks truth with probability $2 p$.
Assertion $(A) :$ If $P$ and $Q$ contradict each other with probability $1 / 2$, then there are two values of $p$.
Reason $(R) :$ A quadratic equation with real coefficients has two real roots.
Answer$(c) :$ Reason is false.
As $P$ and $Q$ contradict each other
$\therefore p(1-2 p)+2 p(1-p)=1 / 2$
$\Rightarrow 8 p^2-6 p+1=0$
$\Rightarrow (2 p-1)(4 p-1)=0$
$\Rightarrow p=1 / 2,1 / 4$
So, assertion is true.
View full question & answer→